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Calculation:

Let $L = \log_{0.5}(\alpha^2)$. Then we know:

$\log_{\alpha^2}(0.5) = \frac{1}{L}$

Given quadratic equation:

$x^2 + Lx + L^4 = 0$

Let roots be $\alpha$ and $\beta$. Then:

$\alpha + \beta = -L$

$\alpha \beta = L^4$

From product of roots: $\beta = \frac{L^4}{\alpha}$

Also given: $\beta^2 = \alpha \cdot \log_{\alpha^2}(0.5) = \frac{\alpha}{L}$

Now equate both expressions for $\beta^2$:

$\left(\frac{L^4}{\alpha}\right)^2 = \frac{\alpha}{L} \Rightarrow \frac{L^8}{\alpha^2} = \frac{\alpha}{L}$

Multiply both sides by $\alpha^2 L$:

$L^9 = \alpha^3 \Rightarrow \alpha = L^3$

Then:

$\beta = \frac{L^4}{\alpha} = \frac{L^4}{L^3} = L$

 $\beta = \log_{0.5}(\alpha^2)$

Hence, the correct answer is Option 2

Calculation:

Let $L = \log_{0.5}(\alpha^2)$. Then the quadratic equation becomes:

$x^2 + Lx + L^4 = 0$

Let the roots be $\alpha$ and $\beta$. Then:

⇒ $\alpha + \beta = -L$

⇒ $\alpha \beta = L^4 \Rightarrow \beta = \frac{L^4}{\alpha}$

Also given: $\beta^2 = \alpha \cdot \log_{\alpha^2}(0.5) = \frac{\alpha}{L}$

Now compute $\beta^2$ from both expressions:

⇒ $\left( \frac{L^4}{\alpha} \right)^2 = \frac{L^8}{\alpha^2}$

Equating both:

⇒ $\frac{L^8}{\alpha^2} = \frac{\alpha}{L} \Rightarrow L^9 = \alpha^3 \Rightarrow \alpha = L^3$

Then:

⇒ $\beta = \frac{L^4}{\alpha} = \frac{L^4}{L^3} = L \Rightarrow \alpha = L^3 = (L)^3 = (\beta)^3 \Rightarrow \alpha = \beta^3$

Now test the given answer $\alpha = 2\beta$:

If $\alpha = 2\beta$, then:

⇒ $\beta = \frac{L^4}{\alpha} = \frac{L^4}{2\beta} \Rightarrow \beta^2 = \frac{L^4}{2}$

Also: $\beta^2 = \frac{\alpha}{L} = \frac{2\beta}{L}$

Equating: $\frac{L^4}{2} = \frac{2\beta}{L} \Rightarrow L^5 = 4\beta \Rightarrow \beta = \frac{L^5}{4}$

Then: $\alpha = 2\beta = \frac{L^5}{2}$ and compare with $\alpha = L^3$

⇒ $L^3 = \frac{L^5}{2} \Rightarrow L^2 = 2 \Rightarrow L = \sqrt{2}$

Then $\alpha = L^3 = 2\sqrt{2}$ and $\beta = \sqrt{2} \Rightarrow \alpha = 2\beta$

Hence, the correct answer is Option 1.

Calculation:

$p = \sum_{j=1}^{n} \log_{10} 2^j \quad \text{and} \quad q = \sum_{j=1}^{n} \log_{10} 5^j$

⇒ $p = \log_{10} 2 \times \sum_{j=1}^{n} j = \log_{10} 2 \times \frac{n(n+1)}{2}$

⇒ $q = \log_{10} 5 \times \frac{n(n+1)}{2}$

⇒ $p + q = \left( \log_{10} 2 + \log_{10} 5 \right) \times \frac{n(n+1)}{2} = \frac{n(n+1)}{2}$

⇒ $p + q = 66 \Rightarrow \frac{n(n+1)}{2} = 66$

⇒ $n(n+1) = 132$

⇒ $n^2 + n - 132 = 0$

⇒ $n = \frac{-1 \pm \sqrt{1^2 - 4(1)(-132)}}{2(1)} = \frac{-1 \pm \sqrt{529}}{2} = \frac{-1 \pm 23}{2}$

⇒  $n = \frac{-1 + 23}{2} = 11 \quad \text{or} \quad n = \frac{-1 - 23}{2} = -12$

Since n  is a positive integer, the value of n is 11.

Thus, the correct answer is:  $9 < n < 12$

Hence, the correct answer is Option 3.

Calculation:

Given:

$p=\sum_{j=1}^{n}\log_{10}2^{,j}$

$=\sum_{j=1}^{n} j\log_{10}2$

$=(1+2+\cdots+n)\log_{10}2 =\frac{n(n+1)}{2}\log_{10}2$

Similarly,

$q=\sum_{j=1}^{n}\log_{10}5^{,j} =\sum_{j=1}^{n} j\log_{10}5$

$=(1+2+\cdots+n)\log_{10}5 =\frac{n(n+1)}{2}\log_{10}5$

Adding $p$ and $q$:

$p+q=\frac{n(n+1)}{2}(\log_{10}2+\log_{10}5) =\frac{n(n+1)}{2}\log_{10}10$

Given $p+q=15$:

$\frac{n(n+1)}{2}\cdot 1=15 \Rightarrow n(n+1)=30$

$\Rightarrow n=5$

Now, find $q-p$:

$q-p=\frac{n(n+1)}{2}(\log_{10}5-\log_{10}2) =\frac{n(n+1)}{2}\log_{10}2.5$

Substitute $n=5$:

$q-p=\frac{5(5+1)}{2}\log_{10}2.5 =15\log_{10}2.5$

Hence, the correct answer is Option 4.

Calculation:

Given, $p = \sin A + \sin B$ and $q = \cos A + \cos B$

We need to evaluate $\frac{p}{q}$.

Using sum-to-product identities:

⇒ $\sin A + \sin B = 2 \sin\left(\tfrac{A + B}{2}\right) \cos\left(\tfrac{A - B}{2}\right)$

⇒ $\cos A + \cos B = 2 \cos\left(\tfrac{A + B}{2}\right) \cos\left(\tfrac{A - B}{2}\right)$

So,

⇒ $\frac{p}{q} = \frac{2 \sin\left(\tfrac{A + B}{2}\right) \cos\left(\tfrac{A - B}{2}\right)}{2 \cos\left(\tfrac{A + B}{2}\right) \cos\left(\tfrac{A - B}{2}\right)}$

Cancel the common factor $2 \cos\left(\tfrac{A - B}{2}\right)$:

⇒ $\frac{p}{q} = \frac{\sin\left(\tfrac{A + B}{2}\right)}{\cos\left(\tfrac{A + B}{2}\right)}$

⇒ $\frac{p}{q} = \tan\left(\tfrac{A + B}{2}\right)$

Hence, the correct answer is Option (c).

Calculation:

We are given the equations:

$\sin A + \sin B = p$ and $\cos A + \cos B = q$

We are asked to find:

$\frac{p^2 - q^2}{p^2 + q^2}$

Using trigonometric identities:

⇒ $\sin(A + B) = \sin A \cos B + \cos A \sin B$

⇒ $\cos(A + B) = \cos A \cos B - \sin A \sin B$

Now, compute $p^2$ and $q^2$:

⇒ $p^2 = (\sin A + \sin B)^2 = \sin^2 A + 2 \sin A \sin B + \sin^2 B$

⇒ $q^2 = (\cos A + \cos B)^2 = \cos^2 A + 2 \cos A \cos B + \cos^2 B$

Next, subtract $q^2$ from $p^2$:

⇒ $p^2 - q^2 = (\sin^2 A + 2 \sin A \sin B + \sin^2 B) - (\cos^2 A + 2 \cos A \cos B + \cos^2 B)$

This simplifies to:

⇒ $p^2 - q^2 = (\sin^2 A - \cos^2 A) + (\sin^2 B - \cos^2 B) + 2 (\sin A \sin B - \cos A \cos B)$

Using the identity $\sin^2 \theta - \cos^2 \theta = -\cos(2\theta)$:

⇒ $p^2 - q^2 = -\cos(2A) - \cos(2B) + 2 \sin(A - B)$

Now, consider the denominator $p^2 + q^2$:

⇒ $p^2 + q^2 = (\sin^2 A + 2 \sin A \sin B + \sin^2 B) + (\cos^2 A + 2 \cos A \cos B + \cos^2 B)$

Which simplifies to:

⇒ $p^2 + q^2 = 2 + 2 \sin(A - B)$

Thus, the expression simplifies to:

⇒ $\frac{p^2 - q^2}{p^2 + q^2} = \cos\left(\frac{\pi}{2} - (A - B)\right)$

Hence, the correct answer is $\cos\left(\frac{\pi}{2} - (A - B)\right)$.

Calculation:

Let $p = \csc 20^\circ = \frac{1}{\sin 20^\circ}$ and $q = \csc 70^\circ = \frac{1}{\sin 70^\circ}$

Note that $\sin 70^\circ = \cos 20^\circ$, so $q = \frac{1}{\cos 20^\circ} = \sec 20^\circ$

We are asked to evaluate:

$\frac{\sqrt{3}p - q}{4} = \frac{\frac{\sqrt{3}}{\sin 20^\circ} - \frac{1}{\cos 20^\circ}}{4}$

Combine the terms:

$= \frac{\sqrt{3} \cos 20^\circ - \sin 20^\circ}{4 \sin 20^\circ \cos 20^\circ}$

Use the identity $\sin 40^\circ = 2 \sin 20^\circ \cos 20^\circ$

Also, $\sin 40^\circ = \sin(60^\circ - 20^\circ) = \sin 60^\circ \cos 20^\circ - \cos 60^\circ \sin 20^\circ$

$= \frac{\sqrt{3}}{2} \cos 20^\circ - \frac{1}{2} \sin 20^\circ$

Multiply both sides by 2:

$2 \sin 40^\circ = \sqrt{3} \cos 20^\circ - \sin 20^\circ$

So the numerator becomes $2 \sin 40^\circ$ and the denominator is $4 \sin 20^\circ \cos 20^\circ = 2 \sin 40^\circ$

Therefore,

$\frac{\sqrt{3}p - q}{4} = \frac{2 \sin 40^\circ}{2 \sin 40^\circ} = 1$

Hence, the correct answer is Option (3): 1.

Calculation:

Given: p = cosec 20° and q = cosec 70°.

The identity for cosecant is: $\csc \theta = \frac{1}{\sin \theta}$

Thus: $p = \frac{1}{\sin 20^\circ}$ and $q = \frac{1}{\sin 70^\circ}$

Now, we compute p² and q²

$p^2 = \left(\frac{1}{\sin 20^\circ}\right)^2 = \frac{1}{\sin^2 20^\circ} \approx 8.554$

$q^2 = \left(\frac{1}{\sin 70^\circ}\right)^2 = \frac{1}{\sin^2 70^\circ} \approx 1.133$

Now, calculate $p^2 + q^2$

$p^2 + q^2 = 8.554 + 1.133 = 9.687$

Also, compute $p^2 q^2$

$p^2 q^2 = 8.554 \times 1.133 = 9.696$

Now, substitute into the given expression:

$\frac{p^2 + q^2}{p^2 q^2} = \frac{9.687}{9.696} \approx 1$

Hence, the correct answer is 1

Calculation:

We are given:

$\cos(2x + 3y) = \frac{1}{2}$ and $\cos(3x + 2y) = \frac{\sqrt{3}}{2}$

Also, $-π < 2x + 3y < π$ and $-π < 3x + 2y < π$

We use known cosine values:

$\cos \theta = \frac{1}{2} \Rightarrow \theta = \pm \frac{π}{3}$

$\cos \theta = \frac{\sqrt{3}}{2} \Rightarrow \theta = \pm \frac{π}{6}$

So possible equations are:

$2x + 3y = \pm \frac{π}{3}$ and $3x + 2y = \pm \frac{π}{6}$

We solve each pair:

Case 1: $2x + 3y = \frac{π}{3},\quad 3x + 2y = \frac{π}{6}$

Solve to get $x = -\frac{π}{30},\quad y = \frac{2π}{15} \Rightarrow x + y = \frac{π}{10}$

Case 2: $2x + 3y = -\frac{π}{3},\quad 3x + 2y = \frac{π}{6}$

Solve to get $x + y = -\frac{π}{2}$

Case 3: $2x + 3y = \frac{π}{3},\quad 3x + 2y = -\frac{π}{6}$

Solve to get $x + y = \frac{π}{2}$

Case 4: $2x + 3y = -\frac{π}{3},\quad 3x + 2y = -\frac{π}{6}$

Solve to get $x + y = -\frac{π}{10}$

So the distinct values of $x + y$ are:

$\frac{π}{10}$, $- π/2$, $\frac{π}{2}$.-π/10

Hence, the correct answer is Option (3): Four.

Calculation:

$\cos(2x+3y)=\frac{1}{2}=\cos\frac{\pi}{3}$

$\therefore; 2x+3y=\frac{\pi}{3} \qquad \text{(i)}$

or $2x+3y=-\frac{\pi}{3}, \quad \left[-\pi<2x+3y<\pi\right] \qquad \text{(ii)}$

Also given:

$\cos(3x+2y)=\frac{\sqrt{3}}{2}=\cos\frac{\pi}{6}$

$\therefore; 3x+2y=\frac{\pi}{6} \qquad \text{(iii)}$

or

$3x+2y=-\frac{\pi}{6}, \quad \left[-\pi<3x+2y<\pi\right] \qquad \text{(iv)}$

Thus we obtain four distinct values of  y−x.

Hence, the correct answer is Option 3.